k( √-2 )
= sinθ
= √2-1
= tan( π/8 )
k’( √-2 )
= cosθ
= √( 2sinθ )
J( √-2 )
= 2^8( k^4-k^2+1 )^3/( k^4-k^2 )^2
= 20^3
a
= ( 2sin2θ )^2
J( ( -1+√-2 )/2 )
= 2^8( a-1 )^3/a
= ( 190-130√2 )^3
k( √-2 )
= sinθ
= √2-1
= tan( π/8 )
k’( √-2 )
= cosθ
= √( 2sinθ )
J( √-2 )
= 2^8( k^4-k^2+1 )^3/( k^4-k^2 )^2
= 20^3
a
= ( 2sin2θ )^2
J( ( -1+√-2 )/2 )
= 2^8( a-1 )^3/a
= ( 190-130√2 )^3
0.
k( 2τ )
= ( 1-k’( τ ) )/( 1+k’( τ ) )
k( τ )
= ( 1-k’( τ/2 ) )/( 1+k’( τ/2 ) )
1.
K( k( 2τ ) )
= ( ( 1+k’( τ ) )/2 )K( k( τ ) )
2.
K’( k( 2τ ) )
= ( 1+k’( τ ) )K’( k( τ ) )
3.
K( k( τ/2 ) = 2√k( τ )/( 1+k( τ ) ) )
= ( 1+k( τ ) )K( k( τ ) )
τ
= ( -1+√-t )/2 ( t > 0 ) :
( k( √-t ) , k’( √-t ) )
= ( sinθ , cosθ )
として
k’( √-t ) = cosθ
= k’( 2τ+1 )
= 1/k’( 2τ )
⇒
k’( 2τ )
= 1/cosθ
⇒
k( 2τ ) = ( 1-k’( τ ) )/( 1+k’( τ ) )
= √-1tanθ
⇒
k’( τ )
= cos2θ-√-1sin2θ
特に
k( τ = ( -1+√-3 )/2 = ζ3 )
= ζ12
⇔
k( ζ3 )^2 = 1-k’( ζ3 )^2 = 1-( cos4θ-√-1sin4θ )
= ζ6
= cos( 2π/6 )+√-1sin( 2π/6 )
⇒
4θ
= 2π/6
⇒
θ
= π/12
⇒
( k( √-3 ) , k’( √-3 ) )
= ( sin( π/12 ) , cos( π/12 ) )
q.e.d.
昨晩久し振りに
園まりさんのCD
を聴いてました!
虫の知らせ?
だったのかなぁ ・・・・・・・
τ
= ( -1+√-t )/2 ( t > 0 ) :
K( k( √-t ) = sinθ )
= ( 1/( 2√cos2θ ) )∫ 1/√( x( 1-x )( 1+( xtan2θ )^2 ) ) dx
( x = 0 〜 1 )
= ( 1/( √cos2θ ) )∫ 1/√( 4u^3-gu-h ) du
( u = 2/3 〜 ∞ )
( g , h > 0 )
g
= 4( 1/3-( tan2θ )^2 )
h
= 4( 2( 1/27+( tan2θ )^2/3 ) )
更に
a
= ( 2sin2θ )^2
として
J( τ )
= 2^8( a-1 )^3/a
= ( 12g )^3/( g^3-27h^2 )
になっている.
τ
= ( -1+√-7 )/2 :
楕円曲線 C : y^2 = 4( u^3-( 20/63 )u-16/189 )
( e1 , e2 , e3 )
= ( 2/3 , -1/3+√( -1/63 ) , -1/3-√( -1/63 ) )
k( τ )
= √( ( e2-e3 )/( e1-e3 ) )
= ( 3+√-7 )/8
J( τ )
= ( -15 )^3
Jを求めなくても
Cは出て来るのか!!
Remark :
g
= 0
⇔
tan2θ
= 1/√3
⇔
θ
= π/12
⇔
τ
= ( -1+√-3 )/2
だから
t
> 3
⇔
g
> 0
K( sin( π/12 ) )
= ( √√3/2 )∫ 1/√( x^3-1 ) dx ( x = 1 〜 ∞ )
= ( √√3/2 )( B( 1/6 , 1/2 )/3 )
= ( √√3/6 )( √3B( 1/3 , 1/2 ) )
= B( 1/3 , 1/2 )/( 2√√3 )
K( sinθ )
= ( 1/cosθ )∫ dx/√( ( 1-x^2 )( 1+( xtanθ )^2 ) )
( x = 0 〜 1 )
= ( 1/( 2√cos2θ ) )∫ dx/√( x( 1-x )( 1+( xtan2θ )^2 ) )
( x = 0 〜 1 )
τ
= ( -1+√-t )/2 ( t > 0 ) :
k( √-t )
= sinθ
k’( √-t )
= cosθ
⇒
k’( τ )
= cos2θ-√-1sin2θ
K( k( τ ) )
= K( sinθ )( cosθ+√-1sinθ ) : 極形式
K( sinθ )
= ∫ 1/√( 4t( 1-tcos2θ )( 1+( tsin2θ )^2 ) ) dt
( t = 0 〜 1/cos2θ )
= ( 1/( 2√cos2θ ) )∫ 1/√( x( 1-x )( 1+( xtan2θ )^2 ) ) dx
( x = 0 〜 1 )
K( 2/√13 )
=
∫ 1/√( 4( x-( 5/13 )x^2 )( 1+( ( 12/13 )x )^2 ) ) dx ( x = 0 〜 13/5 )
( 5/13 )^2+( 12/13 )^2
= 1